Site Help | Search NI Developer Zone DC Motor Calculations, part 1 by Theodore Wildi Electrical Machines, Drives, and Power Systems, Fourth Edition, Prentice Hall PTR Back to Document Now that we have a good understanding of dc generators, we can begin our study of dc motors. Direct-current motors transform electrical energy into mechanical energy. They drive devices such as hoists, fans, pumps, calendars, punch- presses, and cars. These devices may have a definite torque-speed characteristic (such as a pump or fan) or a highly variable one (such as a hoist or automobile). The torque-speed characteristic of the motor must be adapted to the type of the load it has to drive, and this requirement has given rise to three basic types of motors: 1. Shunt motors 2. Series motors 3. Compound motors Direct-current motors are seldom used in ordinary industrial applications because all electric utility systems furnish alternating current. However, for special applications such as in steel mills, mines, and electric trains, it is sometimes advantageous to transform the alternating current into direct current in order to use dc motors. The reason is that the torque-speed characteristics of dc motors can be varied over a wide range while retaining high efficiency. Today, this general statement can be challenged because the availability of sophisticated electronic drives has made it possible to use alternating current motors for variable speed applications. Nevertheless, there are millions of dc motors still in service and thousands more are being produced every year. Table of Contents: � Counter-electromotive force (cemf) � Acceleration of the motor � Mechanical power and torque � Speed of rotation � Armature speed control Direct-current motors are built the same way as generators are; consequently, a dc machine can operate either as a motor or as a generator. To illustrate, consider a dc generator in which the armature, initially at rest, is connected to a dc source Es by means of a switch (Fig. 5.1). The armature has a resistance R, and the magnetic field is created by a set of permanent magnets. As soon as the switch is closed, a large current flows in the armature because its resistance is very low. The individual armature conductors are immediately subjected to a force because they are immersed in the magnetic field created by the permanent magnets. These forces add up to produce a powerful torque, causing the armature to rotate. Figure 5.1 Starting a dc motor across the line. On the other hand, as soon as the armature begins to turn, a second phenomenon takes place: the generator effect. We know that a voltage Eo is induced in the armature conductors as soon as they cut a magnetic field (Fig. 5.2). This is always true, no matter what causes the rotation. The value and polarity of the induced voltage are the same as those obtained Counter-electromotive force (cemf) when the machine operates as a generator. The induced voltage Eo is therefore proportional to the speed of rotation n of the motor and to the flux F per pole, as previously given by Eq. 4.1: Eo = ZnF/60 (4.1) As in the case of a generator, Z is a constant that depends upon the number of turns on the armature and the type of winding. For lap windings Z is equal to the number of armature conductors. In the case of a motor, the induced voltage Eo is called counter-electromotive force (cemf) because its polarity always acts against the source voltage Es. It acts against the voltage in the sense that the net voltage acting in the series circuit of Fig. 5.2 is equal to (Es - Eo) volts and not (Es + Eo) volts. Figure 5.2 Counter-electromotive force (cemf) in a dc motor. The net voltage acting in the armature circuit in Fig. 5.2 is (Es - Eo) volts. The resulting armature current /is limited only by the armature resistance R, and so I = (Es - Eo)IR (5.1) When the motor is at rest, the induced voltage Eo = 0, and so the starting current is I = Es/R The starting current may be 20 to 30 times greater than the nominal full-load current of the motor. In practice, this would cause the fuses to blow or the circuit-breakers to trip. However, if they are absent, the large forces acting on the armature conductors produce a powerful starting torque and a consequent rapid acceleration of the armature. As the speed increases, the counter-emf Eo increases, with the result that the value of (Es — Eo) diminishes. It follows from Eq. 5.1 that the armature current / drops progressively as the speed increases. Although the armature current decreases, the motor continues to accelerate until it reaches a definite, maximum speed. At no-load this speed produces a counter-emf Eo slightly less than the source voltage Es. In effect, if Eo were equal to Es the net voltage (Es — Eo) would become zero and so, too, would the current /. The driving forces would cease to act on the armature conductors, and the mechanical drag imposed by the fan and the bearings would immediately cause the motor to slow down. As the speed decreases the net voltage (Es — Eo) increases and so does the current /. The speed will cease to fall as soon as the torque developed by the armature current is equal to the load torque. Thus, when a motor runs at no- load, the counter-emf must be slightly less than Es so as to enable a small current to flow, sufficient to produce the required torque. Example 5-1 The armature of a permanent-magnet dc generator has a resistance of 1 W and generates a voltage of 50 V when the speed is 500 r/min. If the armature is connected to a source of 150 V, calculate the following: a. The starting current b. The counter-emf when the motor runs at 1000 r/min. At 1460 r/min. c. The armature current at 1000 r/min. At 1460 r/min. Acceleration of the motor Figure 5.3 See Example 5.1. Solution a. At the moment of start-up, the armature is stationary, so Eo = 0 V (Fig. 5.3a). The starting current is limited only by the armature resistance: / = Es/R = 150 V/l W = 150 A b. Because the generator voltage is 50 V at 500 r/min, the cemf of the motor will be 100 V at 1000 r/min and 146 V at 1460 r/min. c. The net voltage in the armature circuit at 1000 r/min is Es - Eo = 150 - 100 = 50 V The corresponding armature current is I = (Es - Eo)/R = 50/1 = 50 A (Fig.5.3b) When the motor speed reaches 1460 r/min, the cemf will be 146 V, almost equal to the source voltage. Under these conditions, the armature current is only / = (Es - Eo)/R = (150 - 146)/1 = 4A and the corresponding motor torque is much smaller than before (Fig. 5.3c). The power and torque of a dc motor are two of its most important properties. We now derive two simple equations that enable us to calculate them. Mechanical power and torque 1. According to Eq. 4.1 the cemf induced in a lap-wound armature is given by Eo = ZnF/60 (4.1) Referring to Fig. 5.2, the electrical power Pa supplied to the armature is equal to the supply voltage Es multiplied by the armature current I: Pa = EsI (5.2) However, Es is equal to the sum of Eo plus the IR drop in the armature: Es = Eo + IR (5.3) It follows that Pa = EsI = (Eo + IR)I = EoI + I2R (5.4) The I2R term represents heat dissipated in the armature, but the very important term EoI is the electrical power that is converted into mechanical power. The mechanical power of the motor is therefore exactly equal to the product of the cemf multiplied by the armature current P = EoI (5.5) where P = mechanical power developed by the motor [W] Eo = induced voltage in the armature (cemf) [V] / = total current supplied to the armature [A] 2. Turning our attention to torque T, we know that the mechanical power P is given by the expression P = nT/9.55 (3.5) where n is the speed of rotation. Combining Eqs. 3.5,4.1, and 5.5, we obtain nT/9.55 = EoI = ZnFI/60 and so T = ZFI/6.28 The torque developed by a lap-wound motor is therefore given by the expression T = ZFI/6.28 (5.6) where T = torque [N×m] Z = number of conductors on the armature F = effective flux per pole [Wb]* / = armature current [A] 6.28 = constant, to take care of units [exact value = 2p] Eq. 5.6 shows that we can raise the torque of a motor either by raising the armature current or by raising the flux created by the poles. Example 5-2 The following details are given on a 225 kW (» 300 hp), 250 V, 1200 r/min dc motor (see Figs. 5.4 and 5.5): armature coils 243 turns per coil 1 type of winding lap armature slots 81 commutator segments 243 field poles 6 diameter of armature 559 mm axial length of armature 235 mm * The effective flux is given by F = 60 Eo/Zn. Figure 5.4 Bare armature and commutator of a dc motor rated 225 kW, 250 V, 1200 r/min. The armature core has a diameter of 559 mm and an axial length of 235 mm. It is composed of 400 stacked laminations 0.56 mm thick. The armature has 81 slots and the commutator has 243 bars. (H. Roberge) Figure 5.5 a. Armature of Fig 5.4 in the process of being wound, coil-forming machine gives the coils the desired shape. b One of the 81 coils ready to be placed in the slots c Connecting the coil ends to the commutator bars. d. Commutator connections ready for brazing (H Roberge) Calculate a. The rated armature current b. The number of conductors per slot c. The flux per pole Solution a. We can assume that the induced voltage Eo is nearly equal to the applied voltage (250 V). The rated armature current is / = P/Eo = 225 000/250 =900A b. Each coil is made up of 2 conductors, so altogether there are 243 X 2 = 486 conductors on the armature. Conductors per slot = 486/81 = 6 Coil sides per slot = 6 c. The motor torque is T = 9.55 P/n = 9.55 X 225 000/1200 = 1791N×m The flux per pole is F = 6.28 T/ZI = (6.28 X 1790)/(486 X 900) = 25.7 mWb When a dc motor drives a load between no-load and full-load, the IR drop due to armature resistance is always small compared to the supply voltage Es. This means that the counter-emf Es is very nearly equal to Es. On the other hand, we have already seen that Eo may be expressed by the equation Eo = ZnF/60 (4.1) Figure 5.6 Ward-Leonard speed control system. Replacing Eo by Es we obtain Es = ZnF/60 That is, where n = speed of rotation [r/min] Es = armature voltage [V] Z = total number of armature conductors This important equation shows that the speed of the motor is directly proportional to the armature supply voltage and inversely proportional to the flux per pole. We will now study how this equation is applied. According to Eq. 5.7, if the flux per pole F is kept constant (permanent magnet field or field with fixed excitation), the speed depends only upon the armature voltage Es. By raising or lowering Es the motor speed will rise and fall in proportion. In practice, we can vary Es by connecting the motor armature M to a separately excited variable-voltage dc generator G (Fig. 5.6). The field excitation of the motor is kept constant, but the generator excitation Ix can be varied from zero to maximum and even reversed. The generator output voltage Es can therefore be varied from zero to maximum, with either positive or negative polarity. Consequently, the motor speed can be varied from zero to maximum in either direction. Note that the generator is driven by an ac motor connected to a 3-phase line. This method of speed control, known as the Ward- Leonard system, is found in steel mills, high-rise elevators, mines, and paper mills. In modem installations the generator is often replaced by a high-power electronic converter that changes the ac power of the electrical utility to dc, by electronic means. The Ward-Leonard system is more than just a simple way of applying a variable dc voltage to the armature of a dc motor. It Speed of rotation Armature speed control can actually force the motor to develop the torque and speed required by the load. For example, suppose Es is adjusted to be slightly higher than the cemf Eo of the motor. Current will then flow in the direction shown in Fig. 5.6, and the motor develops a positive torque. The armature of the motor absorbs power because I flows into the positive terminal. Now, suppose we reduce Es by reducing me generator excitation FG. As soon as Es becomes less than Eo, current/reverses. As a result, (1) the motor torque reverses and (2) the armature of the motor delivers power to generator G. In effect, the dc motor suddenly becomes a generator and generator G suddenly becomes a motor. The electric power that the dc motor now delivers to G is derived at the expense of the kinetic energy of the rapidly decelerating armature and its connected mechanical load. Thus, by reducing Es, the motor is suddenly forced to slow down. What happens to the dc power received by generator G? When G receives electric power, it operates as a motor, driving its own ac motor as an asynchronous generator!* As a result, ac power is fed back into the line that normally feeds the ac motor. The fact that power can be recovered this way makes the Ward-Leonard system very efficient, and constitutes another of its advantages. * The asynchronous generator is explained in Chapter 14. Example 5-3 A 2000 kW, 500 V, variable-speed motor is driven by a 2500 kW generator, using a Ward-Leonard control system shown in Fig. 5.6. The total resistance of the motor and generator armature circuit is 10 mΩ. The motor turns at a nominal speed of 300 r/min, when Eo is 500 V. Calculate a. The motor torque and speed when Es = 400 V and Eo = 380 V b. The motor torque and speed when Es = 350 V and Eo = 380 V Solution a. The armature current is / = (Es - Eo)IR = (400 - 380)/0.01 = 2000 A The power to the motor armature is P = EoI = 380 X 2000 = 760 kW The motor speed is n = (380 V/500 V) X 300 = 228 r/min The motor torque is T = 9.55P/n = (9.55 X 760 000)/228 = 31.8kN×m b. Because Eo = 380 V, the motor speed is still 228 r/min. The armature current is / = (Es - Eo)IR = (350 - 380)/0.01 = -3000 A The current is negative and so it flows in reverse; consequently, the motor torque also reverses. Power returned by the motor to the generator and the 10 mΩ resistance: P = EoI = 380 X 3000 = 1140 kW Braking torque developed by the motor: T = 9.55P/n = (9.55 X 1 140 000)/228 = 47.8 kN×m The speed of the motor and its connected mechanical load will rapidly drop under the influence of this electromechanical braking torque. Rheostat Speed Control Another way to control the speed of a dc motor is to place a rheostat in series with the armature (Fig. 5.7). The current in the rheostat produces a voltage drop which subtracts from the fixed source voltage Es, yielding a smaller supply voltage across the armature. This method enables us to reduce the speed below its nominal speed. It is only recommended for small motors because a lot of power and heat is wasted in the rheostat, and the overall efficiency is low. Furthermore, the speed regulation is poor, even for a fixed setting of the rheostat. In effect, the IR drop across the rheostat increases as the armature current increases. This produces a substantial drop in speed with increasing mechanical load. Figure 5.7 Armature speed control using a rheostat. Related Links: DC Motor Calculations, part 2 DC Motor Calculations, part 3 DC Motor Calculations, part 4 Excerpt from the book published by Prentice Hall PTR. Copyright 2000. Available for purchase online in association with Amazon.com. (Also available for purchase in association with Amazon.co.uk and Amazon.co.de.) My Profile | Privacy | Legal | Contact NI © 2004 National Instruments Corporation. All rights reserved. Site Help | Search NI Developer Zone DC Motor Calculations, part 2 by Theodore Wildi Electrical Machines, Drives, and Power Systems, Fourth Edition, Prentice Hall PTR Back to Document Table of Contents: � Field speed control � Shunt motor under load � Series motor � Series motor speed control � Applications of the series motor According to Eq. 5.7 we can also vary the speed of a dc motor by varying the field flux Φ. Let us now keep the armature voltage Es constant so that the numerator in Eq. 5.7 is constant. Consequently, the motor speed now changes in inverse proportion to the flux Φ if we increase the flux the speed will drop, and vice versa. This method of speed control is frequently used when the motor has to run above its rated speed, called base speed To control the flux (and hence, the speed), we connect a rheostat Rf in series with the field (Fig 5 8a). To understand this method of speed control, suppose that the motor in Fig 5 8a is initially running at constant speed The counter-emf Eo is slightly less than the armature supply voltage Es due to the IR drop in the armature If we suddenly increase the resistance of the rheostat, both the exciting current Ix and the flux Φ will diminish. This immediately reduces the cemf Eo, causing the armature current / to jump to a much higher value. The current changes dramatically because its value depends upon the very small difference between Es and Eo Despite the weaker field, the motor develops a greater torque than before It will accelerate until Eo is again almost equal to Es. Clearly, to develop the same Eo with a weaker flux, the motor must turn faster We can therefore raise the motor speed above its nominal value by introducing a resistance in series with the field For shunt-wound motors, this method of speed control enables high-speed/base-speed ratios as high as 3 to 1. Broader speed ranges tend to produce instability and poor commutation. Under certain abnormal conditions, the flux may drop to dangerously low values For example, if the exciting current of a shunt motor is interrupted accidentally, the only flux remaining is that due to the remanent magnetism in the poles * This flux is so small that the motor has to rotate at a dangerously high speed to induce the required cemf Safety devices are introduced to prevent such runaway conditions. Field speed control Figure 5.8a. Schematic diagram of a shunt motor including the field rheostat b. Torque-speed and torque-current characteristic of a shunt motor. Consider a dc motor running at no-load If a mechanical load is suddenly applied to the shaft, the small no-load current does not produce enough torque to carry the load and the motor begins to slow down This causes the cemf to diminish, resulting in a higher current and a corresponding higher torque When the torque developed by the motor is exactly equal to the torque imposed by the mechanical load, then, and only then, will the speed remain constant (see Section 311) To sum up, as the mechanical load increases, the armature current nses and the speed drops. The speed of a shunt motor stays relatively constant from no-load to full-load In small motors, it only drops by 10 to 15 percent when full-load is applied. In big machines, the drop is even less, due in part, to the very low armature resistance. By adjusting the field rheostat, the speed can, of course, be kept absolutely constant as the load changes. * The term residual magnetism is also used However the IEEE Standard Dictionary of Electrical and Electror ics Terms states ' If there are no air gaps in the mag-netic circuit the remanent induction will equal the residual induction if there are air gaps the remanent induction will be less than the residual induction " Typical torque-speed and torque-current characteristics of a shunt motor are shown in Fig. 5.8b. The speed, torque and current are given in per-unit values. The torque is directly proportional to the armature current. Furthermore, the speed changes only from 1.1 pu to 0.9 pu as the torque increases from 0 pu to 2 pu. Example 5-4 A shunt motor rotating at 1500 r/min is fed by a 120 V source (Fig. 5.9a). The line current is 51 A and the shunt-field resistance is 120 Ω. If the armature resistance is 0.1 Ω, calculate the following; a. The current in the armature b. The counter-emf c. The mechanical power developed by the motor Solution: a. The field current (Fig. 5.9b) is Ix = 120V/120 Ω = 1A The armature current is I = 51 - 1 = 50 A b. The voltage across the armature is E = 120 V Voltage drop due to armature resistance is IR = 50 X 0.1 = 5 V The cemf generated by the armature is Eo = 120 - 5 = 115 V c. The total power supplied to the motor is Pi = EI = 120 X 51 = 6120 W Shunt motor under load Power absorbed by the armature is Pa = EI = 120 X 50 = 6000 W Power dissipated in the armature is P = IR2 = 502 X 0.1 = 250 W Mechanical power developed by the armature is P = 6000 - 250 = 5750 W (equivalent to 5750/746 = 7.7 hp) The actual mechanical output is slightly less than 5750 W because some of the mechanical power is dissipated in bearing friction losses, in windage losses, and in armature iron losses. Figure 5.9 See Example 5.4. A series motor is identical in construction to a shunt motor except for the field. The field is connected in series with the armature and must, therefore, carry the full armature current (Fig. 5.10a). This series field is composed of a few turns of wire having a cross section sufficiently large to carry the current. Although the construction is similar, the properties of a series motor are completely different from those of a shunt motor. In a shunt motor, the flux Φ per pole is constant at all loads because the shunt field is connected to the line. But in a series motor the flux per pole depends upon the armature current and, hence, upon the load. When the current is large, the flux is large and vice versa. Despite these differences, the same basic principles and equations apply to both machines. Figure 5.10a. Series motor connection diagram b. Schematic diagram of a series motor When a series motor operates at full-load, the flux per pole is the same as that of a shunt motor of identical power and speed. However, when the senes motor starts up, the armature current is higher than normal, with the result that the flux per pole is also greater than normal It follows that the starting torque of a senes motor is considerably greater than that of a shunt motor.This can be seen by comparing the T versus / curves of Figs 5. 8 and 5.11. On the other hand, if the motor operates at less than full-load, the armature current and the flux per pole are smaller than normal. The weaker field causes the speed to rise in the same way as it would for a shunt motor with a weak shunt field. For example, if the load current of a senes motor drops to half its normal value, the flux diminishes by half and so the speed doubles. Obviously, if the load is small, the speed may nse to dangerously high values. For this reason we never permit a senes motor to operate at no-load It tends to run away, and the resulting centnfugal forces could tear the windings out of Series motor the armature and destroy the machine When a senes motor carries a load, its speed may have to be adjusted slightly. Thus, the speed can be increased by placing a low resistance in parallel with the series field. The field current is then smaller than before, which produces a drop in flux and an increase in speed. Figure 5.11 Typical speed-torque and current-torque characteristic of a series motor. Conversely, the speed may be lowered by connecting an external resistor in senes with the armature and the field The total IR drop across the resistor and field reduces the armature supply voltage, and so the speed must fall. Typical torque-speed and torque-current characteristics are shown in Fig. 5.11. They are quite different from the shunt motor characteristics given in Fig. 5.8b. Example 5-5 A 15 hp, 240 V, 1780 r/min dc senes motor has a full-load rated current of 54 A. Its operating characteristics are given by the per-unit curves of Fig. 5.11. Calculate a. The current and speed when the load torque is 24 N×m b. The efficiency under these conditions Solution a. We first establish the base power, base speed, and base current of the motor. They correspond to the full-load ratings as follows: PB = 15 hp = 15 X 746 = 11 190 W nB = 1780 r/min /B= 54 A The base torque is, therefore, A load torque of 24 N-m corresponds to a per-unit torque of T(pu) = 24/60 = 0.4 Referring to Fig. 5.11, a torque of 0.4 pu is attained at a speed of 1.4 pu. Thus, the speed is Series motor speed control n = n(pu) X nB = I.4 X 1780 = 2492 r/min From the T vs / curve, a torque of 0.4 pu requires a current of 0.6 pu. Consequently, the load current is / = /(pu) X /B = 0.6 X 54 = 32.4 A b. To calculate the efficiency, we have to know Po and Pi. Pi = EI = 240 X 32.4 = 7776 W Po = nT/9.55 = 2492 X 24/9.55 = 6263 W h = Po/Pi = 6263/7776 = 0.805 or 80.5% Series motors are used on equipment requiring a high starting torque. They are also used to drive devices which must run at high speed at light loads. The series motor is particularly well adapted for traction purposes, such as in electric trains. Acceleration is rapid because the torque is high at low speeds. Furthermore, the series motor automatically slows down as the train goes up a grade yet turns at top speed on flat ground. The power of a series motor tends to be constant, because high torque is accompanied by low speed and vice versa. Series motors are also used in electric cranes and hoists: light loads are lifted quickly and heavy loads more slowly. Related Links: DC Motor Calculations, part 1 DC Motor Calculations, part 3 DC Motor Calculations, part 4 Excerpt from the book published by Prentice Hall PTR. Copyright 2000. Available for purchase online in association with Amazon.com. (Also available for purchase in association with Amazon.co.uk and Amazon.co.de.) Applications of the series motor My Profile | Privacy | Legal | Contact NI © 2004 National Instruments Corporation. All rights reserved. Site Help | Search NI Developer Zone DC Motor Calculations, part 3 by Theodore Wildi Electrical Machines, Drives, and Power Systems, Fourth Edition, Prentice Hall PTR Back to Document Table of Contents: � Compound Motor � Reversing the direction of rotation � Starting a shunt motor � Face-plate starter � Stopping a motor � Dynamic braking A compound dc motor carries both a series field and a shunt field. In a cumulative compound motor, the mmf of the two fields add. The shunt field is always stronger than the series field. Fig. 5.12 shows the connection and schematic diagrams of a compound motor. When the motor runs at no-load, the armature current / in the series winding is low and the mmf of the series field is negligible. However, the shunt field is fully excited by current Ix and so the motor behaves like a shunt machine: it does not tend to run away at no-load. As the load increases, the mmf of the series field increases but the mmf of the shunt field remains constant. The total mmf (and the resulting flux per pole) is therefore greater under load than at no-load. The motor speed falls with increasing load and the speed drop from no-load to full-load is generally between 10 percent and 30 percent. Figure 5.12 a. Connection diagram of a dc compound motor. b. Schematic diagram of the motor. Compound Motor Figure 5.13 Typical speed versus torque characteristics of various dc motors. If the series field is connected so that it opposes the shunt field, we obtain a differential compound motor. In such a motor, the total mmf decreases with increasing load. The speed rises as the load increases, and this may lead to instability. The differential compound motor has very few applications. Fig. 5.13 shows the typical torque-speed curves of shunt, compound and series motors on a per-unit basis. Fig. 5.14 shows a typical application of dc motors in a steel mill. To reverse the direction of rotation of a dc motor, we must reverse either (1) the armature connections or (2) both the shunt and series field connections. The interpoles are considered to form part of the armature. The change in connections is shown in Fig. 5.15. Figure 5.14 Hot strip finishing mill composed of 6 stands, each driven by a 2500 kW dc motor. The wide steel strip is delivered to the runout table (left foreground) driven by 161 dc motors, each rated 3 kW. (Courtesy of General Electric) Reversing the direction of rotation